Name: Physics 52 Refraction and Snell's Law (2 of 11) Light Ray Going Through a Glass Slab
Uploaded: 2024-10-02T09:31:06.000Z
Duration: 11 min 30 s
Description: Thousands of YouTube videos with English-Chinese subtitles! Now you can learn to understand native speakers, expand your vocabulary, and improve your pronunciation...

And here's our next example of how we deal with refraction.

Now this one is a little bit more complicated.

Notice that we have a beam of light traveling through air coming on a boundary between air and glass.

Here's a slab of glass that's 10 centimeters thick.

As it enters the glass, of course glass has a different index of refraction equal to 1.6 as opposed to equal to 1 for air.

You can see that the light will bend, it will refract towards the normal so that the angle of refraction, theta sub 2, is smaller than the angle of incidence, theta sub 1.

Then it travels across the slab, then it reaches the second boundary where you go back from glass into air.

And then it turns out since now the beam is traveling from a region where the index of refraction is larger to a region where the index of refraction is smaller, then the light will now bend away from the normal.

And it turns out since this is air and this is air on both sides of the slab, that the direction of the beam after it leaves the slab will be exactly the same as the direction of the beam as it enters the slab.

The only difference is that because of the refraction, the difference, the change in direction here, there will be an offset if the beam had traveled straight across to where the beam is now.

There will be a difference, a distance, and your job here is to find out what that distance is, how far has that beam been offset by traveling through a 10 centimeter thick slab of glass with an index of refraction of 1.6.

Well the first thing we need to do is find out what theta sub 2 is.

And so therefore we're going to use Snell's Law.

So n1 sine of theta 1 is equal to n2 sine of theta sub 2.

Solving this for theta sub 2, we're going to switch the equation around so we have n2 sine of theta 2 equals n1 sine of theta 1.

Dividing both sides by n2, we get sine of theta 2 equals n1 over n2 times sine of theta 1.

And finally, theta sub 2 is equal to the arc sine, the inverse sine, of n1 over n2 times sine of theta sub 1.

Plugging in the numbers to find out what that is equal to, this is equal to the arc sine of n1, which is 1, divided by n2, which is 1.6, times the sine of theta 1, which was given as 30 degrees.

And so this angle is equal to, here's my calculator, so the sine of 30 is 0.5 divided by 1.6.

And then take the arc sine of that, inverse sine, and we get 18.21 degrees.

I added an extra significant figure so I don't have runoff error at the end.

Now, the ray travels across the glass at a slight angle, so this now will become the angle of incidence of the second boundary, so let's call that theta sub 3.

Well notice that we have two parallel lines here that are bisected by this ray of light right here, and so these two angles here are considered therefore to be what we call alternate, that's what it is, alternate interior angles.

So they're called alternate interior angles and, knowing geometry, those two angles therefore must be equal to each other.

So theta sub 3 is therefore equal to theta sub 2, which therefore is equal to 18.21 degrees.

Then we can see that the beam will then refract outward, and this will then be theta sub 4, and as we said before, theta sub 4 is expected to be the same as theta sub 1, but just so you can see that that's indeed the case, let's go ahead and calculate that.

So here we're going to say that N3 sine of theta sub 3 is equal to N4 sine of theta sub 4.

And so to help us with the index of refractions, notice that we're now going to be working on this boundary right here, so we're going to call this N sub 3, we're going to call this N sub 4, and N sub 3 is equal to 1.6, and N sub 4 is equal to 1, just so we keep things straight.

Coming back over here, we're going to solve for sine of theta sub 4, so we have N4 sine theta sub 4 equals N3 sine of theta sub 3, so we have the sine of theta sub 4 equals N3 over N4 times sine of theta sub 3, and finally, theta sub 4 is equal to the arc sine of N3 over N4 times sine of theta sub 3.

And plug in the numbers, that's equal to the arc sine of N3 is now 1.6, N4 is 1, times the sine of 18.21 degrees.

Let's see what that is equal to, of course we expect that to be 30 degrees, let's find out.

So we take the sine of that, we multiply the times 1.6, and we take the arc sine of that, and sure enough, that's equal to 30 degrees, that's theta sub 4.

That's the exiting angle, the angle of refraction at the second boundary.

Alright, now that we know that, we now need to figure out what D is equal to, so before we start with D, let's figure out what this distance is equal to, right here.

So let's call this distance right here X, and to find what X is equal to, let's draw this line all the way down here, and let's look at this big triangle first, so let's call this X sub, now let's call it X, and then this distance right here, let's call that Y, how about Y?

Just any old variable will do, we'll just call it Y.

So this distance right here is considered X, this distance right there is considered Y.

So I think we can find Y first because we know that this distance here is 10 centimeters, we know that the angle right there between this line and that line is 30 degrees, so let's draw this triangle here on the side, so we have this triangle right here, this line right there, this triangle right here, we know that this here is 30 degrees, we know that this here is 10 centimeters, and this distance right here is equal to Y.

Well we have the opposite side, we have the adjacent side, we have the angle, so we can say that the tangent of 30 degrees is equal to the opposite side, which is Y, over the adjacent side, which is 10 centimeters, so Y is equal to 10 centimeters times the tangent of 30 degrees, and so therefore Y is equal to 30, take the tangent of that, and times 10 equals, and we have 5.77 centimeters.

Alright, so if we know what that is, then we can go ahead and figure out what this distance is, and let's call this distance Z.

I'm going to start running out of variables pretty soon, because then you realize that finally we can say that X is equal to Y minus Z.

Well Z can be found by using a different triangle.

This is now theta sub 2, and theta sub 2 we found to be 18.21 degrees.

We know that this vertical side is still 10 centimeters, and this right here now becomes the side Z that we're trying to find out.

Alright, so let's find Z in this case, we know that the tangent of theta sub 2 is equal to the opposite side, which is Z, divided by the adjacent side, which is 10 centimeters, or Z is equal to 10 centimeters times the tangent of theta sub 2, which is 18.21 degrees.

So Z is equal to, so we have 18.21, take the tangent of that, and multiply that times 10 centimeters, and we have Z to be 3.29 centimeters.

Okay, so now we found Y, we found Z, now we can find X, because X is simply going to be Y minus Z, so let me write that here, X is equal to Y minus Z, and Y is 5.77 centimeters, and Z is 3.29 centimeters, so 5.77 minus 3.29 equals 2.48 centimeters.

We're almost there, so now we have the value for X, so X now is 2.48 centimeters.

Well now we have to make one more triangle.

Let me use a slightly different color here, let's use red, so I'm going to draw a line straight across like this, the length of this red line is equal to D, we know the value for this portion right here, which we called X, and by putting that line there I just kind of destroyed my X, so there's my X.

So we know what X is, we know what D is, and what about this angle right here?

Well this angle has to be equal to this angle, which is equal to 30 degrees, so let me draw this little triangle here on the side, so we have this side here, we have this little line across here, and we have this line across here, and now notice that this line was D, this line was X, and this angle here is 30 degrees, because theta sub 4 is 30 degrees as we found, and this angle must equal this angle.

Well notice that this line is perpendicular to this line, and this line is perpendicular to this line, so the angle between those two lines must equal to the angle between those two lines.

And so finding now what D is equal to, notice that X is the hypotenuse, D is the adjacent side, so we could say that D is equal to the hypotenuse X times the cosine of 30 degrees, because D is adjacent to the angle.

And so finally plugging in what X is equal to right here, which is 2.48 centimeters, so D is equal to 2.48 centimeters times the cosine of 30 degrees, and times 30 cosine equals, and now we found that D is equal to 2.15 centimeters, which is what we're trying to find in the first place.

We had a light beam traveling through a slab that's 10 centimeters thick.

It travels across the first boundary, since you're entering a region that has a high index of refraction, it refracts towards the normal, across the second boundary, since you're now traveling from a high index of refraction region to a low index of refraction region, it bends away from the normal, the exiting ray will be parallel to the entering ray, but they'll be offset by some distance D.

Then you have to find the dimensions of three triangles, this one, the big one, the smaller one, and then this one right here in succession, to find what D is equal to.

 And that's how you use refraction in a case like that.

Subtitles ListPlay Video

Physics 52 Refraction and Snell's Law (2 of 11) Light Ray Going Through a Glass Slab

significant

figure

slightly

recap